The value of is : (a) 0 (b)1/12 (c)1/24 (d)1/64
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First, expand to series t(1+t)/(4+t^4) = t²/4 - (t^3)/8 +O(t^4) Second, integration from t=0 to x leads to I= (x^3)/12 - (x^4)/32 +O(x^5) Then, I/x^3 = 1/12 + x/32 +O(x²) which limit for x -> 0 is 1/12
Originally Posted by sbhatnagar The value of is : (a) 0 (b)1/12 (c)1/24 (d)1/64 Use l'Hospital's Rule: The derivative of the numerator is (using the Fundamental Theorem of Calculs) and the derivative of the denominator is 3x^2. Therefore ....
Thank You! I got it->
Last edited by sbhatnagar; Sep 18th 2011 at 02:02 AM.
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